Finding function given limit












3












$begingroup$


$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?






limits







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  • 1




    $begingroup$
    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    $endgroup$
    – Rebellos
    Nov 23 '18 at 22:45








  • 1




    $begingroup$
    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    $endgroup$
    – JavaMan
    Nov 23 '18 at 22:46
















3












$begingroup$


$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    $endgroup$
    – Rebellos
    Nov 23 '18 at 22:45








  • 1




    $begingroup$
    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    $endgroup$
    – JavaMan
    Nov 23 '18 at 22:46














3












3








3





$begingroup$


$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?






limits







share|cite|improve this question











$endgroup$




$$lim_{xto2} frac{x^2-cx+d}{x^2-4} = 3$$
Find $c$ and $d$.



I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 22:53









gimusi

92.8k84494




92.8k84494










asked Nov 23 '18 at 22:43









Amanda ChoiAmanda Choi

191




191








  • 1




    $begingroup$
    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    $endgroup$
    – Rebellos
    Nov 23 '18 at 22:45








  • 1




    $begingroup$
    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    $endgroup$
    – JavaMan
    Nov 23 '18 at 22:46














  • 1




    $begingroup$
    Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
    $endgroup$
    – Rebellos
    Nov 23 '18 at 22:45








  • 1




    $begingroup$
    Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
    $endgroup$
    – JavaMan
    Nov 23 '18 at 22:46








1




1




$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45






$begingroup$
Hi, Latex/Math-Jax is used to formulate maths in questions and answers, here's a Math-Jax tutorial !. Welcome to our site by the way !
$endgroup$
– Rebellos
Nov 23 '18 at 22:45






1




1




$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46




$begingroup$
Here's where to start: You will need an $x-2$ to cancel from numerator and denominator, so you must have $x^2 - cx + d = 0$ when $x = 2$.
$endgroup$
– JavaMan
Nov 23 '18 at 22:46










2 Answers
2






active

oldest

votes


















4












$begingroup$

Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



$frac{2+a}4=3$ ---> $a=10$



Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



Then $c=-8, d=-20$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    We have that



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



    requires that for $x=2$



    $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



    then



    $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



    then



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



    As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



    $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



    from which we can find $c$ and then $d$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
      $endgroup$
      – Christopher Marley
      Nov 23 '18 at 22:58






    • 1




      $begingroup$
      @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
      $endgroup$
      – gimusi
      Nov 23 '18 at 23:01











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    4












    $begingroup$

    Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



    Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



    $frac{2+a}4=3$ ---> $a=10$



    Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



    Then $c=-8, d=-20$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



      Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



      $frac{2+a}4=3$ ---> $a=10$



      Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



      Then $c=-8, d=-20$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



        Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



        $frac{2+a}4=3$ ---> $a=10$



        Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



        Then $c=-8, d=-20$.






        share|cite|improve this answer











        $endgroup$



        Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.



        Then once we cancel everything out from the original limit, we have $$lim_{xto2} frac{x+a}{x+2}=3$$



        $frac{2+a}4=3$ ---> $a=10$



        Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.



        Then $c=-8, d=-20$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 23 '18 at 22:58

























        answered Nov 23 '18 at 22:50









        Christopher MarleyChristopher Marley

        1,117115




        1,117115























            2












            $begingroup$

            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              $endgroup$
              – Christopher Marley
              Nov 23 '18 at 22:58






            • 1




              $begingroup$
              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              $endgroup$
              – gimusi
              Nov 23 '18 at 23:01
















            2












            $begingroup$

            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              $endgroup$
              – Christopher Marley
              Nov 23 '18 at 22:58






            • 1




              $begingroup$
              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              $endgroup$
              – gimusi
              Nov 23 '18 at 23:01














            2












            2








            2





            $begingroup$

            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.






            share|cite|improve this answer











            $endgroup$



            We have that



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=3$$



            requires that for $x=2$



            $$x^2-cx+d=0 implies 4-2c+d=0 implies d=-4+2c$$



            then



            $$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$



            then



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$



            As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain



            $$lim_{xto 2}frac{x^2-cx+d}{x^2-4}=lim_{xto 2}frac{2x-c}{2x}=3$$



            from which we can find $c$ and then $d$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 '18 at 22:56

























            answered Nov 23 '18 at 22:49









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              $endgroup$
              – Christopher Marley
              Nov 23 '18 at 22:58






            • 1




              $begingroup$
              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              $endgroup$
              – gimusi
              Nov 23 '18 at 23:01


















            • $begingroup$
              Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
              $endgroup$
              – Christopher Marley
              Nov 23 '18 at 22:58






            • 1




              $begingroup$
              @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
              $endgroup$
              – gimusi
              Nov 23 '18 at 23:01
















            $begingroup$
            Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
            $endgroup$
            – Christopher Marley
            Nov 23 '18 at 22:58




            $begingroup$
            Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal.
            $endgroup$
            – Christopher Marley
            Nov 23 '18 at 22:58




            1




            1




            $begingroup$
            @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
            $endgroup$
            – gimusi
            Nov 23 '18 at 23:01




            $begingroup$
            @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result.
            $endgroup$
            – gimusi
            Nov 23 '18 at 23:01


















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